Field of the Invention
The present disclosure relates to an apparatus for modifying voltage command for detecting output current in inverter.
Description of Related Art
In general, a 3-phase inverter serves to drive a 3-phase AC motor in variable frequency, and is variably used due to high power efficiency and enablement of instantaneous torque control. The 3-phase inverter is generally used for control or protection by detecting an output current. Recently, low-priced current detection methods are proposed for household electronics inverters and industrial inverters, and one of the methods is to use a shunt resistor. The output current detection in inverter using the shunt resistor is reasonable in price and low in inverter unit price over the conventional one using a current sensor using Hall effect.
A particular condition must be satisfied in an inverter switching pattern, in order to accurately detect an output current of inverter using the shunt resistor.
FIG. 1 is a circuit diagram illustrating detection of inverter output current using a shunt resistor.
Referring to FIG. 1, the principle of detection of inverter output current using a shunt resistor is such that a DC voltage of DC link capacitor (110) is converted to an AC voltage by a switching unit (120) to be applied to a motor (200), and when a lower switch of the switching unit (120) is turned on, an inverter output current flows to a shunt resistor (130, and an insulation or non-insulation calculation amplifier (140) measures a voltage of the shunt resistor (130) to detect a current.
A current must be sampled to detect the inverter output current, where an inverter voltage command must be modified when the inverter voltage command is in a current detection disabled area.
The switching unit (120) in the 3-phase inverter as in FIG. 1 generally synthesizes a variable size and an AC voltage of variable frequency using a PWM (Pulse Width Modulation) control. Now, a SVPWM (Space Vector PWM) control in the PWM control will be described.
FIG. 2 is an exemplary view illustrating an SVPWM method.
Referring to FIG. 2, a single phase switch includes, as in (a), an upper switch (120a) and a lower switch (120b), which may be equivalently expressed by a switch of (b). (c) and (d) define two switching states, where the state of (c) is defined by Sa=1, the state of (d) is defined by Sa=0.
An output voltage in one switching period includes an ON sequence voltage and an OFF sequence voltage. The ON sequence refers to a stage where a state of upper switch (120a) in one phase is changed from OFF to ON, and the OFF sequence refers to a stage where a state is changed from ON to OFF. When state of each phase is combined, a voltage vector corresponding to a vertex of a hexagon as in FIG. 3 may be generated.
FIG. 3 is an exemplary view illustrating a voltage vector.
Referring to FIG. 3, voltages inside the hexagon may be synthesized using a PWM control. In view of the fact that switching loss can be reduced by minimized state changes of switch, voltage vectors are generally synthesized by combining the switching states corresponding to three vertexes of a triangle including a voltage command.
At this time, a zero vector means a voltage vector where switches on all phases are turned on or off, whereby power is not transmitted due to there being no voltage difference among phases of a motor (200). Furthermore, an effective vector means a voltage vector positioned at a vertex of a hexagon, which is not a zero vector.
FIG. 4 is an exemplary view illustrating synthesis of terminal voltages (Van, Vbn, Vcn) according to PWM method using a triangular wave.
FIG. 4 shows that a switching state is changed from 000(zero vector)→100(effective vector)→110(effective vector)→111(zero vector)→110(effective vector)→100(effective vector)→000, where the switching frequency is minimized because a switch with only one phase is changed when each vector is applied.
FIG. 5 is an exemplary view explaining relationship between FIG. 3 and FIG. 4, where it can be noted that effective times of T1 and T2 may be illustrated in a voltage hexagon of FIG. 3. The voltage hexagon is the output voltage vectors corresponding to the switch states described by the hexagon.
FIG. 6 is an exemplary view illustrating an area where current detection is disabled, and FIG. 7 is an exemplary view illustrating a switching ripple generated from a shunt resistor. Shaded areas in (a) and (b) of FIG. 6 are areas where current detection is disabled.
As in (a), the reason of generating an area in a polygon where current detection is disabled is that, after a switch is turned off as in FIG. 7, a switching ripple is generated on a shunt resistor, and an accurate current can be detected by measurement of voltage only after the switching ripple disappears, which may be expressed by the following Equation 1.Tb=T0+T1=0.5(0.5Tsw−T1−T2)+T1=0.5(0.5Tsw+T1−T2)<Tmin T2−T1>0.5Tsw−2Tmin  [Equation 1]
The Equation 1 is effective when an indirectly calculating method is used in which a sum of 3-phase current is zero where currents of two phases are read and a current of remaining one phase. A condition where a zero vector application time is greater than Tmin must be added in order to obtain a time of Tmin for all 3-phases.
When this condition is added, a current detectable area is reduced in the hexagon, as illustrated in FIG. 6(a). However, in case of a condition for obtaining a time of Tmin for only 2-phases, only three small areas become detection disabled areas as illustrated in FIG. 6(b), whereby a detectable area can be enlarged. Thus, it can be said that a method of reading currents on 2-phases is excellent.
FIG. 8 is an exemplary view illustrating a border between a detection disabled area and a detectable area, where a geometric nature is used in which two sides of an isosceles triangle are same. FIG. 9 is an exemplary view illustrating a time of current sampling, where it can be noted that a current can be sampled when switching states of all phases are zero.
The current detection in the shunt resistor is performed after OFF sequence shunt voltage, and therefore, a current detectable area can be expanded by moving the OFF sequence voltage to a current detectable area, and compensating as much as a difference from the ON sequence voltage, because a current detection on resistance is performed after the OFF sequence voltage.
Voltages are conventionally adjusted by dividing the current detection disabled area to three areas such as S1, S2 and S3.
FIG. 10 is schematic views illustrating a current detection disabled area that is divided into three areas.
That is, injection voltages (Vdss_inj, Vqss_inj) at each area (S1, S2, S3) are calculated, a sum (adjusted voltage) of original voltages and injection voltages is used as a voltage command at OFF sequence, a voltage (compensation voltage) in which an injection voltage is deducted from an original voltage is used as a voltage command at ON sequence, and an average of the ON sequence voltage and OFF sequence voltage is made to be equal to the original voltages (Vdss_org, Vqss_org).
In the conventional method, determination must be made in which area the original voltage command is included, where too many calculations are required to create a burden when a speed of CPU is slow.
Hereinafter, a process of calculating an injection voltage by determining conventional areas will be described.
FIG. 11 is an exemplary view illustrating a process of determining an injection voltage according to prior art.
Referring to FIG. 11, the conventional process of determining an injection voltage is such that a sector is determined (S111), and a voltage command is moved to a sector 1 (S112). The reason of calculation by moving the voltage command to the sector 1 is that other sectors and the sector 1 are on the same position when symmetrically or rotatably moved to allow using the same injection voltage equation.
S111 determines on which sector an original voltage command exists by using a linear Equation of Equation 2. When using a fixed point arithmetic, calculation of √{square root over (3)} may be expressed by the following Equation 3 to thereby require a division calculation. At this time, the number of √{square root over (3)} contained in two linear equations is 2, such that maximum 2 division calculations are required in S111.
                              y          =                                    3                        ⁢            x                          ,                  y          =                                    -                              3                                      ⁢            x                                              [                  Equation          ⁢                                          ⁢          2                ]                                          3                =                  1732          1000                                    [                  Equation          ⁢                                          ⁢          3                ]            
S112 uses Equations 4 to 8. √{square root over (3)} is also contained in this step to require a division calculation. Sector 2 is moved to sector 1 in Equation 4, sector 3 is moved to sector 1 in Equation 5, sector 4 is moved to sector 1 in Equation 6, sector 5 is moved to sector 1 in Equation 7, and sector 6 is moved to sector 1 in Equation 8.
                              [                                                                      V                  ds                  ′                                                                                                      v                  qs                  ′                                                              ]                =                              [                                                                                -                                          1                      2                                                                                                                                  3                                        2                                                                                                                                          3                                        2                                                                                        1                    2                                                                        ]                    ⁡                      [                                                                                V                    ds                    ′                                                                                                                    v                    qs                    ′                                                                        ]                                              [                  Equation          ⁢                                          ⁢          4                ]                                          [                                                                      V                  ds                  ′                                                                                                      v                  qs                  ′                                                              ]                =                              [                                                                                -                                          1                      2                                                                                                                                  3                                        2                                                                                                                    -                                                                  3                                            2                                                                                                            -                                          1                      2                                                                                            ]                    ⁡                      [                                                                                V                    ds                    ′                                                                                                                    v                    qs                    ′                                                                        ]                                              [                  Equation          ⁢                                          ⁢          5                ]                                          [                                                                      V                  ds                  ′                                                                                                      v                  qs                  ′                                                              ]                =                              [                                                                                -                                          1                      2                                                                                                            -                                                                  3                                            2                                                                                                                                        -                                                                  3                                            2                                                                                                            1                    2                                                                        ]                    ⁡                      [                                                                                V                    ds                    ′                                                                                                                    v                    qs                    ′                                                                        ]                                              [                  Equation          ⁢                                          ⁢          6                ]                                          [                                                                      V                  ds                  ′                                                                                                      v                  qs                  ′                                                              ]                =                              [                                                                                -                                          1                      2                                                                                                            -                                                                  3                                            2                                                                                                                                                              3                                        2                                                                                        1                    2                                                                        ]                    ⁡                      [                                                                                V                    ds                    ′                                                                                                                    v                    qs                    ′                                                                        ]                                              [                  Equation          ⁢                                          ⁢          7                ]                                          [                                                                      V                  ds                  ′                                                                                                      v                  qs                  ′                                                              ]                =                              [                                                            1                                                  0                                                                              0                                                                      -                    1                                                                        ]                    ⁡                      [                                                                                V                    ds                    ′                                                                                                                    v                    qs                    ′                                                                        ]                                              [                  Equation          ⁢                                          ⁢          8                ]            
In FIG. 11, an area is determined after moving the voltage command to sector 1 to calculate an injection voltage (S113), and movement is made again to the original sector (S114).
S113 explains a linear equation for dividing the areas. A linear equation for dividing to an area 1 (S1) and a current detectable area may be expressed by the following Equation 9.
                    y        =                                            1                              3                                      ⁢            x                    +                                    2              3                        ⁢                          (                              1                -                                                      4                    ⁢                                                                                  ⁢                                          T                      min                                                                            T                    sw                                                              )                        ⁢                          1                              3                                      ⁢                          V                              d                ⁢                                                                  ⁢                c                                                                        [                  Equation          ⁢                                          ⁢          9                ]            
A line dividing the area 1 (S1) and area 2 (S2) is a straight line FG in FIG. 10, where a point F is a point that internally divides J and D to a ratio of 1:2, and a point G is a point that internally divides A and C to a ratio of 2:3. Obtainment, using the above, of an equation for line FB and an equation for line AC may be expressed by the following Equations 10 and 11.
                    y        =                                                            3                            7                        ⁢            x                    +                                    2              7                        ⁢                          (                              3                -                                                      4                    ⁢                                                                                  ⁢                                          T                      min                                                                            T                    sw                                                              )                        ⁢                          1                              3                                      ⁢                          V                              d                ⁢                                                                  ⁢                c                                                                        [                  Equation          ⁢                                          ⁢          10                ]                                y        =                                            1                              3                                      ⁢            x                    +                                    4              3                        ⁢                          (                              1                -                                                      T                    min                                                        T                    sw                                                              )                        ⁢                          1                              3                                      ⁢                          V                              d                ⁢                                                                  ⁢                c                                                                        [                  Equation          ⁢                                          ⁢          11                ]            
Areas may be determined by using the linear equations, and after the determination, an injection voltage of a relevant area may be determined by the following Equation 12.
                    [                  Equation          ⁢                                          ⁢          12                ]                                                                                  V            dss_inj                    =                                                    -                                  1                  4                                            ⁢                              V                dss_org                ref                                      +                                                            3                                4                            ⁢                              V                qss_org                ref                                      -                                                            V                                      d                    ⁢                                                                                  ⁢                    c                                                  6                            ⁢                              (                                  1                  -                                                            4                      ⁢                                                                                          ⁢                                              T                        min                                                                                    T                      sw                                                                      )                                                    ,                            S        ⁢                                  ⁢        1                                                      V                          q              ⁢              ss_inj                                =                                                                      3                                4                            ⁢                              V                dss_org                ref                                      -                                          3                4                            ⁢                              V                                  q                  ⁢                  ss_org                                ref                                      +                                                                                3                                    ⁢                                      V                                          d                      ⁢                                                                                          ⁢                      c                                                                      6                            ⁢                              (                                  1                  -                                                            4                      ⁢                                                                                          ⁢                                              T                        min                                                                                    T                      sw                                                                      )                                                    ,                            S        ⁢                                  ⁢        1                                                      V            dss_inj                    =                                                    -                                  1                  2                                            ⁢                              V                dss_org                ref                                      +                                                            3                                2                            ⁢                              V                qss_org                ref                                      +                                                            V                                      d                    ⁢                                                                                  ⁢                    c                                                  3                            ⁢                              (                                                      -                    1                                    +                                                            4                      ⁢                                                                                          ⁢                                              T                        min                                                                                    T                      sw                                                                      )                                                    ,                    S2                                                V                          q              ⁢              ss_inj                                =                                                    1                                  2                  ⁢                                      3                                                              ⁢                              V                dss_org                ref                                      -                                          1                2                            ⁢                              V                qss_org                ref                                      -                                                            V                                      d                    ⁢                                                                                  ⁢                    c                                                                    3                  ⁢                                      3                                                              ⁢                              (                                                      -                    1                                    +                                                            4                      ⁢                                                                                          ⁢                                              T                        min                                                                                    T                      sw                                                                      )                                                    ,                    S2                                                V            dss_inj                    =                                    -                              V                dss_org                ref                                      +                                                            V                                      d                    ⁢                                                                                  ⁢                    c                                                  3                            ⁢                              (                                  1                  +                                                            2                      ⁢                                                                                          ⁢                                              T                        min                                                                                    T                      sw                                                                      )                                                    ,                    S3                                                V                          q              ⁢              ss_inj                                =                                    -                              V                                  q                  ⁢                  ss_org                                ref                                      +                                                            V                                      d                    ⁢                                                                                  ⁢                    c                                                                    3                                            ⁢                              (                                  1                  -                                                            2                      ⁢                                                                                          ⁢                                              T                        min                                                                                    T                      sw                                                                      )                                                    ,                    S3      
As noted above, determination of injection voltage may also require the division calculation. The injection voltage is determined at S113, and then the determined voltage is to be converted to an original vector, where the division calculation is required again. Because a voltage in which an original voltage command and the injection voltage are added is dq voltage, the voltage (an original voltage command and the injection voltage are added) is converted to ABC phase voltages, and to PWM command using SVPWM control.
The division calculation at fixed point arithmetic is processed by using several instruction words, and TMS319F2809 (operation clock: 60 MHz) of TI company requires 0.05 μs, for example. The frequency of division for each step according to prior art may be expressed by the following table.
TABLE 1FrequencyCalculation processof divisionSector determination4Area determination11Injection voltage calculation5adjusting/restoring voltage calculation4
As noted above, the conventional current detection method suffers a disadvantage that requires lots of calculations.